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Unit 7
Probability
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 1
WHAT YOU WILL LEARN
• Empirical probability and theoretical
probability
• Compound probability, conditional
probability, and binomial probability
• Odds against an event and odds in
favor of an event
• Expected value
• Tree diagrams
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 2
WHAT YOU WILL LEARN
• Mutually exclusive events and
independent events
• The counting principle,
permutations, and combinations
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 3
Section 4
Expected Value (Expectation)
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 4
Expected Value




E  P1  A1  P2  A2  P3  A3  ...  Pn  An
The symbol P1 represents the probability that
the first event will occur, and A1 represents the
net amount won or lost if the first event occurs.
P2 is the probability of the second event, and A2
is the net amount won or lost if the second
event occurs.
And so on…
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 5
Example

Teresa is taking a multiple-choice test in which
there are four possible answers for each
question. The instructor indicated that she will
be awarded 3 points for each correct answer
and she will lose 1 point for each incorrect
answer and no points will be awarded or
subtracted for answers left blank.
 If Teresa does not know the correct answer to
a question, is it to her advantage or
disadvantage to guess?
 If she can eliminate one of the possible
choices, is it to her advantage or
disadvantage to guess at the answer?
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 6
Solution

Expected value if Teresa guesses.


1
P guesses correctly 
4
3
P guesses incorrectly 
4
1
3
E  3  1
4
4
3 3
  0
4 4


  

Therefore, over the long run, Theresa will neither gain
nor lose points by guessing.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 7
Solution (continued) —eliminate a choice



1
P guesses correctly 
3


2
P guesses incorrectly 
3
1
2
E  3  1
3
3
2 1
 1 
3 3
  

Therefore, over the long run, Theresa will, on average,
gain 1/3 point each time she guesses when she can
eliminate one choice.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 8
Example: Winning a Prize



When Calvin Winters attends a tree farm event,
he is given a free ticket for the $75 door prize. A
total of 150 tickets will be given out. Determine
his expectation of winning the door prize.
 

1
149
E
75 
0
150
150
1

2
Therefore, Calvin’s expectation is $0.50, or 50
cents.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 9
Example

When Calvin Winters attends a tree farm event,
he is given the opportunity to purchase a ticket
for the $75 door prize. The cost of the ticket is
$3, and 150 tickets will be sold. Determine
Calvin’s expectation if he purchases one ticket.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 10
Solution
1
149
E
 72  
 3 
150
150
72 447


150 150
375

150
 2.50

Calvin’s expectation is $2.50 when he
purchases one ticket.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 11
You get to select one card at random from a
standard deck of 52 cards. If you pick a king, you
win $6. If you pick a queen, you lose $3 and if
you pick a jack, you lose $2. Determine your
expectation for this game.
a.
$0.08
b.
$0.46
c.
$0.77
d.
$1.00
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 12
You get to select one card at random from a
standard deck of 52 cards. If you pick a king, you
win $6. If you pick a queen, you lose $3 and if
you pick a jack, you lose $2. Determine your
expectation for this game.
a.
$0.08
b.
$0.46
c.
$0.77
d.
$1.00
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 13
Fair Price
Fair price = expected value + cost to play
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 14
Example



Suppose you are playing a game in which you
spin the pointer shown in the figure, and you are
awarded the amount shown under the pointer. If
it costs $10 to play the game, determine:
a) the expectation of the
$10
$2
person who plays the
$15
$20
game.
$10
$2
b) the fair price to play the
game.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 15
Solution

Amt. Shown
on Wheel
$2
$10
$15
$20
Probability
3/8
3/8
1/8
1/8
Amount
Won/Lost
$8
$0
$5
$10
       
3
3
1
1
E  $8  $0  $5  $10
8
8
8
8
24
5 10

0 
8
8 8
9

 1.125  $1.13
8
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 16
Solution

Fair price = expectation + cost to play
= $1.13 + $10
= $8.87
Thus, the fair price is about $8.87.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 17
Section 5
Tree Diagrams
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 18
Counting Principle

If a first experiment can be performed in M
distinct ways and a second experiment can be
performed in N distinct ways, then the two
experiments in that specific order can be
performed in M • N distinct ways.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 19
Definitions

Sample space: A list of all possible outcomes
of an experiment.

Sample point: Each individual outcome in the
sample space.

Tree diagrams are helpful in determining
sample spaces.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 20
Example

a)
b)
c)
d)
Two balls are to be selected without
replacement from a bag that contains one
purple, one blue, and one green ball.
Use the counting principle to determine the
number of points in the sample space.
Construct a tree diagram and list the sample
space.
Find the probability that one blue ball is
selected.
Find the probability that a purple ball followed
by a green ball is selected.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 21
Solutions
a) 3 • 2 = 6 ways
b)
B PB
P
G PG
BP
B
P
G
G
P
BG
GP
B
GB
Copyright © 2009 Pearson Education, Inc.


c)
4 2
P blue  
6 3
d)
P Purple,Green

1
P P,G
6

Chapter 12 Section 1 - Slide 22
One die is rolled and one colored chip - black or
white - is selected at random. Use the counting
principle to determine the number of sample points
in the sample space.
a.
6
b.
8
c.
12
d.
10
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 23
One die is rolled and one colored chip - black or
white - is selected at random. Use the counting
principle to determine the number of sample points
in the sample space.
a.
6
b.
8
c.
12
d.
10
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 24
P(event happening at least once)
 event happening
 event does 
P
 1 P 

 at least once

 not happen
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 25
Section 6
Or and And Problems
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 26
Or Problems


P(A or B) = P(A) + P(B)  P(A and B)
Example: Each of the numbers 1, 2, 3, 4, 5, 6,
7, 8, 9, and 10 is written on a separate piece of
paper. The 10 pieces of paper are then placed
in a bowl and one is randomly selected. Find
the probability that the piece of paper selected
contains an even number or a number greater
than 5.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 27
Solution

P(A or B) = P(A) + P(B)  P(A and B)
 even or

P


 greater than 5
 even and

P even  P greater than 5  P 
 greater than 5

 

5
5
3
7




10 10 10 10

Thus, the probability of selecting an even number or a
number greater than 5 is 7/10.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 28
Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and
10 is written on a separate piece of paper. The
10 pieces of paper are then placed in a bowl
and one is randomly selected. Find the
probability that the piece of paper selected
contains a number less than 3 or a number
greater than 7.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 29
Solution


2
P less than 3 
10


3
P greater than 7 
10
There are no numbers that are both less than 3
and greater than 7. Therefore,
 less than 3 or  2
3
5 1
P


0


10 2
 greater than 7 10 10
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 30
Mutually Exclusive

Two events A and B are mutually exclusive if
it is impossible for both events to occur
simultaneously.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 31
Example

One card is selected from a standard deck of
playing cards. Determine the probability of the
following events.
a) selecting a 3 or a jack
b) selecting a jack or a heart
c) selecting a picture card or a red card
d) selecting a red card or a black card
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 32
Solutions
a) 3 or a jack
(mutually exclusive)
 

4
4
P 3  P jack 

52 52
8
2


52 13
b) jack or a heart
 jack and
4 13 1
P jack  P heart  P 




 heart
 52 52 52

 

16 4


52 13
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 33
Solutions continued
c) picture card or red card
 picture &  12 26 6
P picture  P red  P 




 red card  52 52 52

  
d) red card or black card
(mutually exclusive)
  

26 26
P red  P black 

52 52
52

1
52
Copyright © 2009 Pearson Education, Inc.
32 8


52 13
Chapter 12 Section 1 - Slide 34
And Problems


P(A and B) = P(A) • P(B)
Example: Two cards are to be selected with
replacement from a deck of cards. Find the
probability that two red cards will be selected.
     
P A  P B  P red  P red
26 26


52 52
1 1 1
  
2 2 4
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 35
Example

Two cards are to be selected without
replacement from a deck of cards. Find the
probability that two red cards will be selected.
P  A   P  B   P  red  P  red
26 25


52 51
1 25 25
 

2 51 102
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 36
Independent Events

Event A and Event B are independent events
if the occurrence of either event in no way
affects the probability of the occurrence of the
other event.

Experiments done with replacement will result in
independent events, and those done without
replacement will result in dependent events.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 37
One die is rolled and one colored chip - black or
white - is selected at random. Determine the
probability of obtaining an even number and the
color white.
a.
1
2
c.
1
6
Copyright © 2009 Pearson Education, Inc.
b.
1
3
d.
1
4
Slide 12 - 38
One die is rolled and one colored chip - black or
white - is selected at random. Determine the
probability of obtaining an even number and the
color white.
a.
1
2
c.
1
6
Copyright © 2009 Pearson Education, Inc.
b.
1
3
d.
1
4
Slide 12 - 39
Example

A package of 30 tulip bulbs contains 14 bulbs
for red flowers, 10 for yellow flowers, and 6 for
pink flowers. Three bulbs are randomly selected
and planted. Find the probability of each of the
following.
a.All three bulbs will produce pink flowers.
b.The first bulb selected will produce a red
flower, the second will produce a yellow
flower and the third will produce a red flower.
c. None of the bulbs will produce a yellow
flower.
d.At least one will produce yellow flowers.
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 40
Solution

30 tulip bulbs, 14 bulbs for red flowers,
10 for yellow flowers, and 6 for pink flowers.
a. All three bulbs will produce pink flowers.

 
 
 
P 3 pink  P pink 1  P pink 2  P pink 3

6 5 4



30 29 28
1

203
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 41
Solution (continued)
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.
b. The first bulb selected will produce a red flower,
the second will produce a yellow flower and the
third will produce a red flower.


   
  
P red, yellow, red  P red  P yellow  P red
14 10 13



30 29 28
13

174
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 42
Solution (continued)
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.

c. None of the bulbs will produce a yellow flower.
 none 
 first not   second not   third not 
P
 P
P 
P 



 yellow
 yellow   yellow
  yellow 
20 19 18



30 29 28
57

203
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 43
Solution (continued)

30 tulip bulbs, 14 bulbs for red flowers,
10 for yellow flowers, and 6 for pink flowers.
d. At least one will produce yellow flowers.
P(at least one yellow) = 1  P(no yellow)
57
 1
203
146

203
Copyright © 2009 Pearson Education, Inc.
Chapter 12 Section 1 - Slide 44